Question

Solve the initial-value problem $$ y^{\prime}=x^{2}-4 x+3-6 e^{x}, \quad y(0)=8 . $$

Short answer

The solution is \(y(x) = \frac{x^3}{3} - 2x^2 + 3x - 6e^x + 14\).

Step-by-step solution

Identify the Problem

We are given a differential equation \(y' = x^2 - 4x + 3 - 6e^x\) with an initial condition \(y(0) = 8\). Our goal is to find the function \(y(x)\) that satisfies both the differential equation and the initial condition.

Integrate the Differential Equation

To find \(y(x)\), we need to integrate the right-hand side of the differential equation. The integral of \(y'\) is \(y\), so we will integrate \(x^2 - 4x + 3 - 6e^x\) with respect to \(x\).

Perform the Integration

- Integrate \(x^2\): \(\int x^2 \, dx = \frac{x^3}{3}\).- Integrate \(-4x\): \(\int -4x \, dx = -2x^2\).- Integrate \(3\): \(\int 3 \, dx = 3x\).- Integrate \(-6e^x\): \(\int -6e^x \, dx = -6e^x\).Putting these together, we get:\[y(x) = \frac{x^3}{3} - 2x^2 + 3x - 6e^x + C\] where \(C\) is the constant of integration.

Apply the Initial Condition

We use the initial condition \(y(0) = 8\) to solve for \(C\). Substitute \(x = 0\) and \(y = 8\) into the equation:\[8 = \frac{0^3}{3} - 2(0)^2 + 3(0) - 6e^0 + C\]This simplifies to:\[8 = -6 + C\]Solving for \(C\) gives us: \(C = 14\).

Write the Final Solution

Substitute \(C = 14\) back into the equation for \(y(x)\):\[y(x) = \frac{x^3}{3} - 2x^2 + 3x - 6e^x + 14\]This is the function that satisfies both the differential equation and the initial condition.

Key concepts

Understanding Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In this context, a derivative represents how a function changes as its input changes. When you see a differential equation like \(y' = x^2 - 4x + 3 - 6e^x\), it is telling you about the rate of change of the function \(y\) in terms of \(x\) at any point. The "prime" symbol (') in \(y'\) is shorthand for the derivative of \(y\). Differential equations can describe many kinds of real-world phenomena, from the movement of particles to the spread of diseases. They are a powerful tool in engineering, physics, economics, and beyond. When solving differential equations, the goal typically is to find a function \(y(x)\) that satisfies both the equation itself and any given initial or boundary conditions.

The Process of Integration

Integration is the process of finding the function \(y(x)\) whose derivative is known. It can be thought of as the inverse operation to differentiation. In essence, differentiation gives us the slope of a function, while integration helps us find the original function when its slope is known. When you integrate the expression \(x^2 - 4x + 3 - 6e^x\), you break it down into simpler parts:

• \(\int x^2 \, dx = \frac{x^3}{3}\)
• \(\int -4x \, dx = -2x^2\)
• \(\int 3 \, dx = 3x\)
• \(\int -6e^x \, dx = -6e^x\)

These individual integrations are combined to form the integrated function, plus a constant \(C\) which we determine later through initial conditions. Integration plays a vital role in solving many mathematical problems, particularly those involving areas, volumes, and accumulated quantities.

Applying Initial Conditions

Initial conditions provide specific values for the function and its derivatives at a certain point, which helps in finding the particular solution to a differential equation. Think of them as clues that help us pinpoint the exact form of the solution among the many possibilities.In this problem, the initial condition is \(y(0) = 8\). This tells us that when \(x = 0\), the value of the function \(y\) must be 8. We use this information to solve for the constant \(C\) found during the integration step. By plugging \(x = 0\) into our integrated function \(y(x) = \frac{x^3}{3} - 2x^2 + 3x - 6e^x + C\) and setting it equal to 8, we find \(C = 14\). Once \(C\) is determined, we have a complete function \(y(x)\) that satisfies both the differential equation and the initial condition. This specific solution is crucial for accurately modeling or predicting real-world scenarios that the equation represents.